A ball is dropped from the top of a building that is 784 feet high. The height of the ball above ground level is given by the polynomial function h(t)= -16t to the second power + vot + ho where t is measured in seconds, vo is the initial velocity and ho is the initial height.
What is the formula after all the known values are put in?
How high is the ball after 3 seconds?
How far has the ball traveled in 3 seconds?
When will the ball hit the ground?
h(t) = -16t^2 + 784
because v0 = 0
Now just plug in your numbers as needed. If you get stuck, show how far you got, and we can help out.
1040
The formula for the height of the ball above ground level is h(t) = -16t^2 + vot + ho. Given that the building height is 784 feet, the initial height ho would be 784 feet.
To find the height of the ball after 3 seconds, we substitute t=3 into the formula:
h(3) = -16(3)^2 + vo(3) + 784
h(3) = -144 + 3vo + 784
h(3) = 3vo + 640
To find how far the ball has traveled in 3 seconds, we need to find the horizontal distance. The formula for horizontal distance (d) is d = vo*t, where vo is the initial velocity and t is the time. As the ball is dropped, the initial horizontal velocity is 0, so vo = 0. Therefore, the horizontal distance traveled in 3 seconds would be 0.
To find when the ball will hit the ground, we need to find the time when h(t) = 0, since at that moment, the height above the ground is 0. Let's solve for t in the equation -16t^2 + vot + ho = 0:
-16t^2 + vot + ho = 0
-16t^2 + 0t + 784 = 0
Divide by -16: t^2 - 49 = 0
(t - 7)(t + 7) = 0
So, the ball will hit the ground at t = 7 seconds or t = -7 seconds. Since time cannot be negative in this context, the ball will hit the ground at t = 7 seconds.
To find the formula after plugging in the known values, we need to substitute the given values into the polynomial function:
h(t) = -16t^2 + vot + ho
Given:
Initial height (ho) = 784 feet
Initial velocity (vo) = 0 (since the ball is dropped, the initial velocity is zero)
Plugging in these values, the formula becomes:
h(t) = -16t^2 + 0t + 784
= -16t^2 + 784
To find the height of the ball after 3 seconds, we need to substitute t = 3 into the formula:
h(3) = -16(3)^2 + 784
= -16(9) + 784
= -144 + 784
= 640 feet
Therefore, the ball is at a height of 640 feet after 3 seconds.
To find how far the ball has traveled in 3 seconds, we need to calculate the horizontal distance. Since the ball is dropped vertically with no initial velocity in the horizontal direction, it doesn't move horizontally. Therefore, the ball hasn't traveled any horizontal distance and will be at the same horizontal position as where it was dropped.
To find when the ball will hit the ground, we need to set h(t) = 0 and solve for t. When the ball hits the ground, its height will be zero.
0 = -16t^2 + 784
Rearranging the equation:
16t^2 = 784
Dividing both sides by 16:
t^2 = 49
Taking the square root of both sides:
t = ±√49
t = ±7
Since time cannot be negative, the ball will hit the ground after 7 seconds.